You appear to be avoiding symmetries. Exploit them, instead.
I didn't look at it real closely, but are you SURE you need [-pi/2,pi/2] or will twice the result from [0,pi/2] do the trick?
Hey, all. Here is the initial problem:
What it the volume of the solid inside both the regions:
I set the integral up as follows:
But that doesn't work for some reason, because taking θ from -π/2 to π/2 cancels out, and instead, we have to take θ from π to 2π, like so:
I don't understand why that is.
EDIT: I also don't understand why my LaTeX notation isn't working. Not even simple **** like .
EDIT #2: Is it perhaps because both r and θ need to be positive when finding a volume? If so, why would that be true for volume but not area?
The problem isn't finding the answer, but knowing why one way works and the other doesn't.
The final answer, by the way, is:
Well, if things cancel out, this suggests you are crossing an important boundary, like f(x) = x over [-1,1]. What happens at x = 0?
There are a couple of things that seem to be overlooked.
(a^2)^(3/2) = |a| <> a
We really need a little more information about 'a'.
You've the same problem with the sine function.
Over [-pi/2,pi/2], the sine certainly changes sign, so it DOES make a difference.
Lastly, your third integral in the iteration contains the term [sin(x)]^3.
This is EXACTLY what I was talkling about in first sentence.
sin(x) is an odd function.
So also [sin(x)]^3.
Note: You may also wish to get constants out of integrals and don't let them confuse you. You don't need the 'r' until the second one. You don't need the '2' until you are done.